Answer : The energy of one photon of hydrogen atom is, ![3.03\times 10^{-19}J](https://tex.z-dn.net/?f=3.03%5Ctimes%2010%5E%7B-19%7DJ)
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:
![\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%5Cleft%28%5Cfrac%7B1%7D%7Bn_i%5E2%7D-%5Cfrac%7B1%7D%7Bn_f%5E2%7D%20%5Cright%20%29)
Where,
= Wavelength of radiation
= Rydberg's Constant = 10973731.6 m⁻¹
= Higher energy level = 3
= Lower energy level = 2
Putting the values, in above equation, we get:
![\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3D%2810973731.6%29%5Cleft%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%20%5Cright%20%29)
![\lambda=6.56\times 10^{-7}m](https://tex.z-dn.net/?f=%5Clambda%3D6.56%5Ctimes%2010%5E%7B-7%7Dm)
Now we have to calculate the energy.
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
h = Planck's constant = ![6.626\times 10^{-34}Js](https://tex.z-dn.net/?f=6.626%5Ctimes%2010%5E%7B-34%7DJs)
c = speed of light = ![3\times 10^8m/s](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E8m%2Fs)
= wavelength = ![6.56\times 10^{-7}m](https://tex.z-dn.net/?f=6.56%5Ctimes%2010%5E%7B-7%7Dm)
Putting the values, in this formula, we get:
![E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%286.626%5Ctimes%2010%5E%7B-34%7DJs%29%5Ctimes%20%283%5Ctimes%2010%5E8m%2Fs%29%7D%7B6.56%5Ctimes%2010%5E%7B-7%7Dm%7D)
![E=3.03\times 10^{-19}J](https://tex.z-dn.net/?f=E%3D3.03%5Ctimes%2010%5E%7B-19%7DJ)
Therefore, the energy of one photon of hydrogen atom is, ![3.03\times 10^{-19}J](https://tex.z-dn.net/?f=3.03%5Ctimes%2010%5E%7B-19%7DJ)
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Answer:
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